(a) Average time to get to school
Average time (minutes) = Summation of the two means = mean time to walk to bus stop + mean time for the bust to get to school = 8+20 = 28 minutes
(b) Standard deviation of the whole trip to school
Standard deviation for the whole trip = Sqrt (Summation of variances)
Variance = Standard deviation ^2
Therefore,
Standard deviation for the whole trip = Sqrt (2^2+4^2) = Sqrt (20) = 4.47 minutes
(c) Probability that it will take more than 30 minutes to get to school
P(x>30) = 1-P(x=30)
Z(x=30) = (mean-30)/SD = (28-30)/4.47 ≈ -0.45
Now, P(x=30) = P(Z=-0.45) = 0.3264
Therefore,
P(X>30) = 1-P(X=30) = 1-0.3264 = 0.6736 = 67.36%
With actual average time to walk to the bus stop being 10 minutes;
(d) Average time to get to school
Actual average time to get to school = 10+20 = 30 minutes
(e) Standard deviation to get to school
Actual standard deviation = Previous standard deviation = 4.47 minutes. This is due to the fact that there are no changes with individual standard deviations.
(f) Probability that it will take more than 30 minutes to get to school
Z(x=30) = (mean - 30)/Sd = (30-30)/4.47 = 0/4.47 = 0
From Z table, P(x=30) = 0.5
And therefore, P(x>30) = 1- P(X=30) = 1- P(Z=0.0) = 1-0.5 = 0.5 = 50%
Answer:
D. no solution
Step-by-step explanation:
h-z=3(1)
h=3+z(minus z from both side)
put in the second equation
-3h,+3z=6(2)
-3(3+z)+3z=6
-9-3z+3z=6
-9=6( not true)
so the equation doesn't have a solution.
